Ado's theorem states that given a finite-dimensional Lie algebra $\mathfrak g$, there exists a faithful representation $\rho\colon\mathfrak g \to \mathfrak{gl}(V)$, with $V$ a finite-dimensional vector space. In the real or complex case one can take the exponent of the image and obtain a (virtual) Lie subgroup $\exp\rho(\mathfrak g)$ in $GL(V)$ having Lie algebra $\rho(\mathfrak g)$. But nothing guarantees that this subgroup will be closed in $GL(V)$.

So the question is: is every finite-dimensional Lie algebra the Lie algebra of some closed linear Lie group? I am primarily interested in the real and complex case, but it might be interesting to ask what happens in the ultrametric case as well.

3 Answers
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I think that the answer is yes. It looks like you can prove it by relying on a convenient proof of Ado's theorem.

Procesi's book, "Lie groups: an approach through invariants and representations", has the following theorem preceding the proof of Ado's theorem:

Theorem 2. Given a Lie algebra $L$ with semismiple part $A$, we can embed it into a new Lie algebra $L'$ with the following properties:

$L'$ has the same semismiple part $A$ as $L$.

The solvable radical of $L'$ is decomposed as $B' \oplus N'$, where $N'$ is the nilpotent radical of $L'$, $B'$ is an abelian Lie algebra acting by semisimple derivations, and $[A, B'] = 0$.

$A \oplus B'$ is a subalgebra and $L' = (A \oplus B') \ltimes N'$.

With all of that, the idea is to first prove the refinement of Ado's theorem for $L'$. We need a particular refinement: Let $\tilde{A}$ be the maximal algebraic semisimple Lie group with Lie algebra $A$, and let $\tilde{B'}$ and $\tilde{N'}$ be the contractible Lie groups with Lie algebras $B'$ and $N'$. If we can find a closed embedding of $(\tilde{A} \times \tilde{B'}) \ltimes \tilde{N'}$ in a matrix group, then it will restrict to a closed embedding of the Lie subgroup of the original $L$.

In the proof of Ado's theorem that follows, the action of $N'$ is nilpotent, so the representation of $\tilde{N'}$ is closed and faithful. The Lie algebra $L'$ has a representation which is trivial on $B'$ and $N'$ and generates $\tilde{A}$. It has another representation which is trivial on $N'$ and $A$ and for which the action of $B'$ is nilpotent. If I have not made a mistake, the direct sum of these three representations is the desired representation of $L'$.

Greg, thank you for your answer! I didn't understand few things: 1. Why "then it will restrict to a closed embedding of the Lie subgroup of the original L."? 2. If we take direct sum of the three rep's, it looks like ~N' will normalize ~A x ~B', am I wrong? 3. Do we work over C here? It looks like the proof of Theorem 2 is given for complex case only. Is it crucial?
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mathreaderDec 3 '09 at 4:46

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1. $L$ generates a closed subgroup $\tilde{L}$ of $\tilde{L'}$, and closed within closed is closed. 2. I don't understand what you're saying. The first summand is the most important one; in this summand $\tilde{N'}$ certainly does not normalize the other factors. 3. For the same reason that you can pass to $L'$, you can also pass to the complexification, so it should still work over the reals.
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Greg KuperbergDec 3 '09 at 4:54

Consider the simplest case: the irrational winding of two-dimensional torus. Here $\mathfrak{g}$ is one-dimensional real Lie algebra isomorphic to $\mathbb{R}$. The representation is given to us by $\rho: \mathfrak{g}\to \diag(i\mathbb{R},i\mathbb{R})$, $x\mapsto \diag(ix, i\alpha x)$, where $\alpha$ is irrational. Then $\exp\rho(\mathfrak{g})$ is a non-closed, or virtual, Lie subgroup of $U(1)\times U(1)$.
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mathreaderJul 24 '11 at 11:28